GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Dec 2018, 03:13

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### 10 Keys to nail DS and CR questions

December 17, 2018

December 17, 2018

06:00 PM PST

07:00 PM PST

Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong.
• ### R1 Admission Decisions: Estimated Decision Timelines and Chat Links for Major BSchools

December 17, 2018

December 17, 2018

10:00 PM PST

11:00 PM PST

From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.

# If p+q = j and p-q = k, then 2pq =

Author Message
TAGS:

### Hide Tags

Manager
Joined: 26 Apr 2010
Posts: 108
Concentration: Strategy, Entrepreneurship
Schools: Fuqua '14 (M)
If p+q = j and p-q = k, then 2pq =  [#permalink]

### Show Tags

26 Nov 2010, 12:11
2
7
00:00

Difficulty:

(N/A)

Question Stats:

76% (01:50) correct 24% (02:38) wrong based on 329 sessions

### HideShow timer Statistics

If p+q = j and p-q = k, then 2pq =

A. (j^2 - k^2)/2
B. (k^2 - j^2)/2
C. jk/2
D. (j^2 + k^2)/2
E. (j - k)/2

Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

Can someone post the algebraic solution? Thanks in advance.

_________________

I appreciate the kudos if you find this post helpful! +1

Math Expert
Joined: 02 Sep 2009
Posts: 51259

### Show Tags

26 Nov 2010, 12:21
4
martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.

Given: $$p+q = j$$ and $$p-q = k$$ --> square both expressions: $$p^2+2pq+q^2=j^2$$ and $$p^2-2pq+q^2=k^2$$ --> now, subtract (2) form (1): $$4pq=j^2-k^2$$ --> $$2pq=\frac{j^2-k^2}{2}$$.

_________________
Manager
Joined: 26 Apr 2010
Posts: 108
Concentration: Strategy, Entrepreneurship
Schools: Fuqua '14 (M)

### Show Tags

26 Nov 2010, 12:43
Thanks very much Bunuel.

+1
_________________

I appreciate the kudos if you find this post helpful! +1

Manager
Joined: 13 Jul 2010
Posts: 136

### Show Tags

26 Nov 2010, 13:03
Bunuel wrote:
martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.

Given: $$p+q = j$$ and $$p-q = k$$ --> square both expressions: $$p^2+2pq+q^2=j^2$$ and $$p^2-2pq+q^2=k^2$$ --> now, subtract (2) form (1): $$4pq=j^2-k^2$$ --> $$2pq=\frac{j^2-k^2}{2}$$.

Bunuel did you know to square all of the terms once you looked at the answers - since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach.
Manager
Joined: 26 Apr 2010
Posts: 108
Concentration: Strategy, Entrepreneurship
Schools: Fuqua '14 (M)

### Show Tags

26 Nov 2010, 13:23
Good question...my thought exactly...I didn't 'see' that I should square both sides...
_________________

I appreciate the kudos if you find this post helpful! +1

Intern
Joined: 03 Jun 2009
Posts: 45

### Show Tags

27 Nov 2010, 09:49
gettinit wrote:
Bunuel wrote:
martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.

Given: $$p+q = j$$ and $$p-q = k$$ --> square both expressions: $$p^2+2pq+q^2=j^2$$ and $$p^2-2pq+q^2=k^2$$ --> now, subtract (2) form (1): $$4pq=j^2-k^2$$ --> $$2pq=\frac{j^2-k^2}{2}$$.

Bunuel did you know to square all of the terms once you looked at the answers - since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach.

These kind of questions wud b easier if we make a note that....

(a+b)^2 = a^2 + b^2 + 2ab
(a-b)^2 = a^2 + b^2 -2ab

=> a^2 + b^2 = [(a+b)^2 + (a-b)^2]/2...Sum of the squares *0.5

also

=> 4ab = [(a+b)^2 - (a-b)^2]/2 ...difference of the squares *0.5
Manager
Joined: 01 Nov 2010
Posts: 126
Location: Zürich, Switzerland

### Show Tags

Updated on: 29 Nov 2010, 15:22
Look right at 2pq which hints at use of the formula- (x+y)2 and (x-y)2

subtracting (x-y)2 from (x+y)2 above gives 4pq which when divided by 2 gives the result.

Originally posted by Sarang on 27 Nov 2010, 17:46.
Last edited by Sarang on 29 Nov 2010, 15:22, edited 1 time in total.
Manager
Joined: 30 Aug 2010
Posts: 88
Location: Bangalore, India

### Show Tags

29 Nov 2010, 06:27
1
martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.

MAKE IT SIMPLE.

==> 2p=j+k
SUBTRACT 2nd equation from 1st
==> 2q=j-k

multiply the above 2 equations
4pq=j^2-K^2
==> 2pq = j^2-K^2 / 2

Regards,
Murali.

Kudos?
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If p+q = j and p-q = k, then 2pq =  [#permalink]

### Show Tags

17 Jul 2014, 01:53
p+q = j ........... (1)
p-q = k ............ (2)

(1) + (2)

$$p = \frac{j+k}{2}$$

(1) - (2)

$$q = \frac{j-k}{2}$$

$$2pq = 2 * \frac{j+k}{2} *\frac{j-k}{2}$$

$$= \frac{j^2 - k^2}{2}$$

_________________

Kindly press "+1 Kudos" to appreciate

Intern
Joined: 14 Aug 2016
Posts: 1
Location: India
WE: Consulting (Consulting)
Re: If p+q = j and p-q = k, then 2pq =  [#permalink]

### Show Tags

26 Oct 2016, 23:06
1
martie11 wrote:
If p+q = j and p-q = k, then 2pq =

A. (j^2 - k^2)/2
B. (k^2 - j^2)/2
C. jk/2
D. (j^2 + k^2)/2
E. (j - k)/2

Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

Can someone post the algebraic solution? Thanks in advance.

here you go,

(p+q)^2 = p^2 + q^2 +2pq = j^2 ................1
(p-q)^2 = p^2 + q^2 -2pq = k^2 ..................2

subtract 2 from 1

4pq=j^2 - k^2 ........................3

divide eqaution 3 by 2 on both sides

2pq = (j^2 - k^2)/2

which is option A.
Non-Human User
Joined: 09 Sep 2013
Posts: 9197
Re: If p+q = j and p-q = k, then 2pq =  [#permalink]

### Show Tags

29 Jun 2018, 21:45
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If p+q = j and p-q = k, then 2pq = &nbs [#permalink] 29 Jun 2018, 21:45
Display posts from previous: Sort by